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zysi [14]
3 years ago
9

An eagle carrying a trout flies above a lake along a horizontal path. The eagle drops the trout from a height of 6.1 m. The fish

travels 7.9 m horizontally before hitting the water. What is the velocity of the eagle? Round your answer to the nearest tenth.
Physics
2 answers:
VikaD [51]3 years ago
8 0
The answer is 7.1 I just took the test
ololo11 [35]3 years ago
8 0
 7.1 I just took the test
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What do you do abt onion eyes (MY EYES ARE BURNING BADDD)
MissTica

Answer:

Leave onions in cold water for about 15 minutes! Takes out the chemical reaction in the onion's defense system.

Explanation:

This is what people NEED to know for cooking... Lol :)

8 0
2 years ago
Which of the following does NOT represent Newton’s second law? Question 20 options: a = m/Fnet m = Fnet/a Fnet = ma a = Fnet/m
Natali [406]

Answer:

a=m/f is not an equation under newton's second law

Explanation:

newton's second law of motion is represented using: f=ma

where a=v-u/t

therefore it becomes,f=m(v-u)/t

from f=ma,

a will become f/m,

m will become f/a

8 0
3 years ago
A ray is incident on a film of thickness t and the index of refraction n=1.6 at an angle θ = 35 degrees. Find the angle of refra
lidiya [134]

Answer

Given,

refractive index of film, n = 1.6

refractive index of air, n' = 1

angle of incidence, i = 35°

angle of refraction, r = ?

Using Snell's law

n' sin i = n sin r

1 x sin 35° = 1.6 x sin  r

 r = 21°

Angle of refraction is equal to 21°.

Now,

distance at which refractive angle comes out

d = 2.5 mm

α be the angle with horizontal surface and incident ray.

α = 90°-21° = 69°

t be the thickness of the film.

So,

tan \alpha = \dfrac{t}{d/2}

tan 61^0 = \dfrac{t}{2.5/2}

t = 2.26 mm

Hence, the thickness of the film is equal to 2.26 mm.

4 0
3 years ago
A 3.00-kg object has a velocity 1 6.00 i ^ 2 2.00 j ^2 m/s. (a) what is its kinetic energy at this moment? (b) what is the net w
tatyana61 [14]
(a) The velocity of the object on the x-axis is 6 m/s, while on the y-axis is 2 m/s, so the magnitude of its velocity is the resultant of the velocities on the two axes:
v= \sqrt{(6.00m/s)^2+(2.00 m/s)^2}=6.32 m/s
And so, the kinetic energy of the object is
K= \frac{1}{2}mv^2= \frac{1}{2}(3.00 kg)(6.32 m/s)^2=60 J

(b) The new velocity is 8.00 m/s on the x-axis and 4.00 m/s on the y-axis, so the magnitude of the new velocity is
v= \sqrt{(8.00 m/s)^2+(4.00 m/s)^2}=8.94 m/s
And so the new kinetic energy is
K= \frac{1}{2}mv^2= \frac{1}{2}(3.00 kg)(8.94 m/s)^2=120 J

So, the work done on the object is the variation of kinetic energy of the object:
W=\Delta K=120 J-60 J=60 J
7 0
3 years ago
Who was the first person to describe the earth as a magnet
denis-greek [22]

Answer:

William Gilbert

Explanation:

first described the Earth as a giant dipole magnet 400 years ago. But, as Rod Wilson recounts, he did far more than this.

5 0
3 years ago
Read 2 more answers
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